Problem 19 - Differential Equations and Linear Algebra

Question

Determine the inverse Laplace transform of \(F.\) $$F(s)=\frac{e^{-2 s}}{s^{2}+2 s+2}$$.

Step-by-Step Solution

Answer

The inverse Laplace transform of \(F(s) = \frac{e^{-2 s}}{s^{2}+2 s+2}\) is \(f(t) = -\cos(1(t-2))u(t-2)\).

Step 1: Identify the known Laplace transform pair

We first need to identify an already known Laplace transform pair in the given function. Observe that the denominator can be rewritten as: \(s^2 + 2s + 2 = (s+1)^2 + 1\) Now, notice that the function appears to resemble the Laplace transform of cosine: If \(\mathcal{L}\{-\cos(at)\} = \frac{s}{s^2 + a^2}\), then comparing the denominators, we can write the function as: \(F(s) = \frac{e^{-2s}}{s^{2}+2 s+2} = \frac{e^{-2s}}{(s+1)^2 + 1^2}\)

Step 2: Apply the Laplace transform property

The given function has an exponential term \(e^{-2s}\) in the numerator. We can use the property of the Laplace transform to shift the result: \(\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a)\), where a=2. Now apply the inverse Laplace transform while considering the shift. \(\mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{e^{-2s} \cdot \frac{1}{(s+1)^2 + 1^2}\}\)

Step 3: Inverse Laplace transform

Now, we'll apply the inverse Laplace transform to the function: \(f(t) = \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{e^{-2s} \cdot \frac{1}{(s+1)^2 + 1^2}\}\) Using the property discussed above and recognizing the Laplace transform pair, we have: \(f(t) = (f(t-2)\cdot u(t-2))\) Where \(f(t-2)\) is the inverse Laplace transform of \(\frac{1}{(s+1)^2 + 1^2}\), which is a cosine function with frequency 1 and a shift of -1 in the s-axis: \(f(t-2) = -\cos(1(t-2))\)

Step 4: Final answer

Combining the functions gives the inverse Laplace transform for the given function: \(f(t) = -\cos(1(t-2))u(t-2)\) So, the inverse Laplace transform of \(F(s) = \frac{e^{-2 s}}{s^{2}+2 s+2}\) is: \(f(t) = -\cos(1(t-2))u(t-2)\)

Key Concepts

Laplace Transform Pair

When studying the Laplace transform, a fundamental concept is the idea of a Laplace transform pair. This involves understanding that each function in the time domain correlates to a unique function in the s-domain, and vice versa. The Laplace transform takes a time-domain function, often denoted as f(t), and transforms it into an s-domain function F(s).

For practical applications, including solving differential equations, it's useful to recognize these pairs. For instance, common transform pairs include the Laplace transform of a sine function and its counterpart, the exponential function in the s-domain. Recognizing these pairs allows one to quickly determine the inverse Laplace transform, which is the process of converting back from the s-domain to the time domain. If you know that a particular form of F(s) corresponds to a known function f(t), you can essentially 'look up' the inverse without performing complex integrations every time.

Furthermore, in the context of our problem, knowing that the denominator \(s^2 + 2s + 2 = (s+1)^2 + 1\) shares similarity to the denominator in the Laplace transform of a cosine function \(\frac{s}{s^2 + a^2}\), where \(a\) is the angular frequency, enables us to link them as a transform pair, seamlessly shifting our perspective from the frequency to the time domain.

Exponential Shift Property

The exponential shift property is another powerful tool in the analysis of Laplace transforms. It refers to how an exponential multiplier in the s-domain affects the time-domain function. Specifically, if you encounter an expression of the form \(e^{-as}F(s)\), it indicates a time delay in the original function by \(a\) units. In the time domain, this translates to a shifted function \(f(t-a)\) and is often accompanied by a unit step function \(u(t-a)\) to ensure the shifted function starts at \(t = a\).

In the exercise, we encountered such an expression: \(e^{-2s}\frac{1}{(s+1)^2 + 1^2}\). Here, \(e^{-2s}\) serves as a 'signal' that the time-domain equivalent will be delayed by 2 seconds, turning on at \(t = 2\). This understanding of the exponential shift property is critical for correctly reconstructing the time-domain signal from its transformed counterpart when working with inverse Laplace transforms.

Unit Step Function

The unit step function, denoted as \(u(t-a)\), is a pivotal function in time-domain analysis, serving as a mathematical 'switch.' It effectively turns on a function at a certain point, \(a\text{.}\) Before this point, the function is multiplied by zero (the switch is off), and after this point, it is multiplied by one (the switch is on).

This concept is especially important when solving differential equations or analyzing systems with delayed responses. In the provided exercise, the function \(u(t-2)\) activates the cos function, \(f(t)\), starting from \(t = 2\). This models a system response that doesn't begin until 2 seconds into the observation period. Understanding how the unit step function operates is crucial when interpreting inverse Laplace transforms in a piecewise manner, offering a clear depiction of time-shifted behaviors in signal and system analysis.